Given, $\overrightarrow{F}=\left(t\hat{\mathrm{i}}+3t^2\hat{\mathrm{j}}\right) \mathrm{N}$ and $m=1 \mathrm{kg}$

Now, we can write from Newton's second law, $\overrightarrow{F}=m\overrightarrow{a}=m\frac{d\overrightarrow{v}}{dt}$

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

Integrating the above, 

$$\begin{gathered} {\int }_0^{v}d\overrightarrow{v}={\int }_0^t\left(t\hat{\mathrm{i}}+3t^2\hat{\mathrm{j}}\right)dt \\ \Rightarrow \overrightarrow{v}=\frac{t^2}{2}\hat{\mathrm{i}}+t^3\hat{\mathrm{j}} \end{gathered}$$

Now, power developed by the force is $P=\overrightarrow{F}·\overrightarrow{v}=\left(t\hat{\mathrm{i}}+3t^2\hat{\mathrm{j}}\right)·\left(\frac{t^2}{2}\hat{\mathrm{i}}+t^3\hat{\mathrm{j}}\right)$

$=\left(\frac{t^3}{2}+3t^5\right)$

At $t=2 \mathrm{s}$, power is $P=\left(\frac{2^3}{2}+3{\left(2\right)}^5\right)=4+\left(3\times 32\right)=100 \mathrm{W}$