Solution

First, we prove the well‑known formula for the field pressure, which will be convenient to use in other problems:
$$P=\frac{dF}{dS}=E\frac{dq}{dS}=\sigma E_{external}.$$
For a conductor in vacuum, the field near its surface is related to the surface charge density by:
$$\sigma=\varepsilon_0 E.$$
This is easily obtained from Gauss's theorem, remembering that inside the conductor the field is zero.

Immediately near the surface, an area element behaves like an infinite plane. Such a plane creates its own field of $\sigma/(2\varepsilon_0)$ on each side. Then the external field is
$$E_{external}=\frac{\sigma}{\varepsilon_0}-\frac{\sigma}{2\varepsilon_0}=\frac{E}{2}.$$
From this we finally get:
$$\boxed{p=\frac{\varepsilon_0 E^2}{2}}\tag{1}$$

In problem 6.4.5 the capacitance of a spherical capacitor is found:
$$C=4\pi\varepsilon_0\frac{Rr}{R-r}\tag{2}$$
Accordingly, the charge on each sphere is
$$Q=VC=4\pi\varepsilon_0V\frac{Rr}{R-r}.$$
The field near the inner sphere is
$$E=\frac{Q}{4\pi\varepsilon_0r^2}=V\frac{R}{r(R-r)}\tag{3}$$
$(3)\to(1)$
$$P=\frac{\varepsilon_0R^2V^2}{2r^2(R-r)^2}.$$