Answer
$B'=\{ (\frac{1}{\sqrt{5}},\frac{2}{\sqrt{5}}),(-\frac{2}{\sqrt{5}},\frac{1}{\sqrt{5}})\}$ is orthonormal basis for $R^2$.
Work Step by Step
Let $B=\{ v_1=(1,2),v_2=(-1,0) \}$.
Applying the Gram-Schmidt orthonormalization process produces
$$w_1=v_1=(1,2)$$
$$w_2=v_2- \frac{v_2\cdot w_1}{w_1\cdot w_1}w_1=(-1,0)+\frac{1}{5}(1,2)=(-\frac{4}{5},\frac{2}{5}).$$
Normalizing $w_1$and $w_2$ produces the orthonormal set
$$u_1=\frac{w_1}{\|w_1\|}=(\frac{1}{\sqrt{5}},\frac{2}{\sqrt{5}})$$
$$u_2=\frac{w_2}{\|w_2\|}=\frac{\sqrt{5}}{2}(-\frac{4}{5},\frac{2}{5})=
(-\frac{2}{\sqrt{5}},\frac{1}{\sqrt{5}}).$$
Hence, $B'=\{ (\frac{1}{\sqrt{5}},\frac{2}{\sqrt{5}}),(-\frac{2}{\sqrt{5}},\frac{1}{\sqrt{5}})\}$ is orthonormal basis for $R^2$.