The equation for velocity is 

$\overrightarrow{v}=\left(2t \hat{\mathrm{i}} + 3t^2 \hat{\mathrm{j}}\right)$

Thus, the acceleration is 

$\overrightarrow{a}=\frac{d\overrightarrow{v}}{dt}=\frac{d(2t \hat{\mathrm{i}}+3t^2 \hat{\mathrm{j}})}{dt}=\left(2 \hat{\mathrm{i}}+6t \hat{\mathrm{j}}\right)$

At $t=1 \mathrm{s}$, $\overrightarrow{a}=2 \hat{\mathrm{i}} + 6 \hat{\mathrm{j}}$

The force is written as $\overrightarrow{F}=m\overrightarrow{a}=\frac{2 \hat{\mathrm{i}}+6 \hat{\mathrm{j}}}{2}=\hat{\mathrm{i}} +3 \hat{\mathrm{j}}$

Comparing it with the given value of force, $x=3$.