Answer
$${\bf{N}}\left( 2 \right) = - \frac{{2\sqrt 5 }}{5}{\bf{i}} + \frac{{\sqrt 5 }}{5}{\bf{j}}$$
Work Step by Step
$$\eqalign{
& {\bf{r}}\left( t \right) = t{\bf{i}} + \frac{1}{2}{t^2}{\bf{j}},{\text{ }}t = 2 \cr
& {\text{Calculate }}{\bf{r}}{\text{'}}\left( t \right) \cr
& {\bf{r}}'\left( t \right) = \frac{d}{{dt}}\left[ {t{\bf{i}} + \frac{1}{2}{t^2}{\bf{j}}} \right] \cr
& {\bf{r}}'\left( t \right) = {\bf{i}} + t{\bf{j}} \cr
& {\text{Find the unit Tangent Vector }}{\bf{T}}\left( t \right) \cr
& {\bf{T}}\left( t \right) = \frac{{{\bf{r}}'\left( t \right)}}{{\left\| {{\bf{r}}'\left( t \right)} \right\|}},{\text{ }}{\bf{r}}'\left( t \right) \ne 0 \cr
& {\bf{T}}\left( t \right) = \frac{{{\bf{i}} + t{\bf{j}}}}{{\left\| {{\bf{i}} + t{\bf{j}}} \right\|}} = \frac{{{\bf{i}} + t{\bf{j}}}}{{\sqrt {1 + {t^2}} }} \cr
& {\text{Find }}{\bf{T}}'\left( t \right) \cr
& {\bf{T}}'\left( t \right) = \frac{d}{{dt}}\left[ {\frac{1}{{\sqrt {1 + {t^2}} }}{\bf{i}} + \frac{t}{{\sqrt {1 + {t^2}} }}{\bf{j}}} \right] \cr
& {\bf{T}}'\left( t \right) = \frac{d}{{dt}}\left[ {{{\left( {1 + {t^2}} \right)}^{ - 1/2}}{\bf{i}} + t{{\left( {1 + {t^2}} \right)}^{ - 1/2}}{\bf{j}}} \right] \cr
& {\bf{T}}'\left( t \right) = \frac{d}{{dt}}\left[ {{{\left( {1 + {t^2}} \right)}^{ - 1/2}}} \right]{\bf{i}} + \underbrace {\frac{d}{{dt}}\left[ {t{{\left( {1 + {t^2}} \right)}^{ - 1/2}}} \right]}_{{\text{Product rule}}}{\bf{j}} \cr
& {\bf{T}}'\left( t \right) = - \frac{1}{2}{\left( {1 + {t^2}} \right)^{ - 3/2}}\left( {2t} \right){\bf{i}} + \left[ { - \frac{1}{2}t{{\left( {1 + {t^2}} \right)}^{ - 3/2}}\left( {2t} \right) + {{\left( {1 + {t^2}} \right)}^{ - 1/2}}} \right]{\bf{j}} \cr
& {\bf{T}}'\left( t \right) = - \frac{t}{{{{\left( {1 + {t^2}} \right)}^{3/2}}}}{\bf{i}} + \left[ { - {t^2}{{\left( {1 + {t^2}} \right)}^{ - 3/2}} + {{\left( {1 + {t^2}} \right)}^{ - 1/2}}} \right]{\bf{j}} \cr
& {\bf{T}}'\left( t \right) = - \frac{t}{{{{\left( {1 + {t^2}} \right)}^{3/2}}}}{\bf{i}} + {\left( {1 + {t^2}} \right)^{ - 3/2}}\left[ { - {t^2} + 1 + {t^2}} \right]{\bf{j}} \cr
& {\bf{T}}'\left( t \right) = - \frac{t}{{{{\left( {1 + {t^2}} \right)}^{3/2}}}}{\bf{i}} + \frac{1}{{{{\left( {1 + {t^2}} \right)}^{3/2}}}}{\bf{j}} \cr
& {\text{Evaluate }}{\bf{T}}'\left( t \right){\text{ at }}t = 2 \cr
& {\bf{T}}'\left( 2 \right) = - \frac{2}{{{{\left( {1 + {2^2}} \right)}^{3/2}}}}{\bf{i}} + \frac{1}{{{{\left( {1 + {2^2}} \right)}^{3/2}}}}{\bf{j}} \cr
& {\bf{T}}'\left( 2 \right) = - \frac{2}{{{5^{3/2}}}}{\bf{i}} + \frac{1}{{{5^{3/2}}}}{\bf{j}} \cr
& \left\| {{\bf{T}}'\left( 2 \right)} \right\| = \sqrt {{{\left( { - \frac{2}{{{5^{3/2}}}}} \right)}^2} + {{\left( {\frac{1}{{{5^{3/2}}}}} \right)}^2}} = \sqrt {\frac{4}{{125}} + \frac{1}{{125}}} = \frac{1}{5} \cr
& {\text{Finding the Principal Unit Normal Vector}} \cr
& {\bf{N}}\left( t \right) = \frac{{{\bf{T}}'\left( t \right)}}{{\left\| {{\bf{T}}'\left( t \right)} \right\|}} \cr
& {\text{Evaluate }}{\bf{N}}'\left( t \right){\text{ at }}t = 2 \cr
& {\bf{N}}\left( 2 \right) = \frac{{{\bf{T}}'\left( 2 \right)}}{{\left\| {{\bf{T}}'\left( 2 \right)} \right\|}} = \frac{{ - \frac{2}{{{5^{3/2}}}}{\bf{i}} + \frac{1}{{{5^{3/2}}}}{\bf{j}}}}{{1/5}} \cr
& {\bf{N}}\left( 2 \right) = - \frac{{2\left( 5 \right)}}{{{5^{3/2}}}}{\bf{i}} + \frac{5}{{{5^{3/2}}}}{\bf{j}} \cr
& {\bf{N}}\left( 2 \right) = - \frac{2}{{{5^{1/2}}}}{\bf{i}} + \frac{1}{{{5^{1/2}}}}{\bf{j}} \cr
& {\bf{N}}\left( 2 \right) = - \frac{2}{{\sqrt 5 }}{\bf{i}} + \frac{1}{{\sqrt 5 }}{\bf{j}} \cr
& {\bf{N}}\left( 2 \right) = - \frac{{2\sqrt 5 }}{5}{\bf{i}} + \frac{{\sqrt 5 }}{5}{\bf{j}} \cr} $$
