Given, mass of object, $m_2=2 \mathrm{~kg}$
Velocity, $\mathbf{v}=\left(8 t \hat{\mathbf{i}}+3 t^2 \hat{\mathbf{j}}\right) \mathrm{m} / \mathrm{s}$
$\therefore$ Acceleration of object is given as
$\begin{aligned}
& \mathbf{a}=\frac{d \mathbf{v}}{d t}=\frac{d}{d t}\left(8 t \hat{\mathbf{i}}+3 t^2 \hat{\mathbf{j}}\right) \\
& \mathbf{a}=8 \hat{\mathbf{i}}+6 t \hat{\mathbf{j}} \mathrm{ms}^{-2}
\end{aligned}$
Hence, according to Newton's second law of motion, force on the object,
$\mathbf{F}=m \mathbf{a}=2(8 \hat{\mathbf{i}}+6 t \hat{\mathbf{j}})$


$\begin{array}{rlrl}
& |\mathbf{F}| =\sqrt{16^2+(12 t)^2} \\
\Rightarrow & 20 \left.=\sqrt{16^2+(12 t)^2} \quad \text { [Given, }|\mathbf{F}|=20 \mathrm{~N}\right] \\
\Rightarrow & 400 =256+144 t^2 \\
\Rightarrow & 144 t^2 =144 \Rightarrow t^2=1 \quad \Rightarrow t=1 \mathrm{~s}
\end{array}$
Put $t=1 \mathrm{~s}$ in Eq. (i), we get
$\mathbf{F}=16 \hat{\mathbf{i}}+12 \hat{\mathbf{j}}$
$\therefore$ Angle made by force $\mathbf{F}$ relative to positive direction of $X$-axis is given as
$\theta=\tan ^{-1}\left(\frac{F_Y}{F_X}\right)=\tan ^{-1}\left(\frac{12}{16}\right)=\tan ^{-1}\left(\frac{3}{4}\right)$