Answer
$$k = \frac{{\ln 7}}{2}$$
Work Step by Step
$$\eqalign{
& y = {e^{2x}} \cr
& {\text{The area is given by}} \cr
& A = \int_0^k {{e^{2x}}} dx \cr
& {\text{Let }}A = 3 \cr
& 3 = \int_0^k {{e^{2x}}} dx \cr
& {\text{Integrating}} \cr
& 3 = \frac{1}{2}\left[ {{e^{2x}}} \right]_0^k \cr
& 3 = \frac{1}{2}\left[ {{e^{2k}} - {e^0}} \right] \cr
& {\text{Solve for }}k \cr
& 3 = \frac{1}{2}{e^{2k}} - \frac{1}{2} \cr
& 6 = {e^{2k}} - 1 \cr
& {e^{2k}} = 7 \cr
& \ln {e^{2k}} = \ln 7 \cr
& 2k = \ln 7 \cr
& k = \frac{{\ln 7}}{2} \cr} $$