Solution

Let's consider the following figure

From right figure

$$\cos{\theta} = \frac{R}{2R} = 0.5$$

so, $\theta = 60^\circ$. Applying Newton Second Law for submerged trunk on $x$-axis,

$$N = N_t~\sin{\theta} \tag{1}$$

on $y$-axis

$$F_{A1} = mg + N_t~\cos{\theta}$$

$$\rho_w~g~V = mg + N_t~\cos{\theta} \tag{2}$$

For half-submerged trunk, on $x$-axis,

$$N = N_t~\sin{\theta}$$

on $y$-axis

$$F_{A2} + N_t~\cos{\theta} = mg$$

$$\rho_w~g~\frac{V}{2} + N_t~\cos{\theta} = mg \tag{3}$$

Substituting $(2)$ into $(3)$, it is obtained

$$N_t = \frac{mg}{3~\cos{\theta}} \tag{4}$$

Putting $(4)$ into $(1)$

$$N = \frac{mg}{3}~\tan{\theta}$$

$$\boxed{N = \frac{mg}{\sqrt{3}}}$$

Note: You can prove that both pressure forces generated by trunks are equals.