Torque due to the force $F$ on a thin spherical shell,
$$
\begin{aligned}
\tau & =r \times F \\
& =2 R F \sin 90^{\circ}=2 R F\left[\because \sin 90^{\circ}=1\right]
\end{aligned}
$$

Where, $I$ is moment of inertia of a thin spherical shell.

From parallel axes's theorem, moment of inertia of spherical shell,
$$
I=I_{\mathrm{cm}}+M r^2
$$
or
$$
I=\frac{2}{3} M R^2+M R^2 \quad(\because r=R)
$$

From Eqs. (i), we get
$$
\begin{gathered}
\alpha=\frac{2 R F}{\frac{2}{3} M R^2+M R^2} \\
=\left(\frac{6}{5}\right) \frac{F}{R M}
\end{gathered}
$$

Hence, the tangential acceleration, $a_T=R \alpha$ or
$$
a_T=\left(\frac{6}{5}\right) \frac{F}{R M} \times R=\frac{6}{5} \frac{F}{M}
$$