Solution

Essentially, we need to find the charge at which the electric field pressure becomes equal to $P$.

The surface charge density is
$$\sigma = \frac{\rho}{2\pi R} \tag{1}$$
where $\rho = \frac{dq}{dl}$ is the desired quantity.

Near the surface of the cylinder, the field is
$$E = \frac{\sigma}{\varepsilon_0} \tag{2}$$
This is trivially derived from Gauss's theorem.

In the previous problem, the formula for the field pressure was proved (it is, incidentally, equal to the energy density of the field):
$$P = \frac{\varepsilon_0 E^2}{2} \tag{3}$$
$(1)\to(2)\to(3):$
$$P = \frac{\rho^2}{8\varepsilon_0 \pi^2 R^2} \tag{3}$$
$$\rho = 2\pi R \sqrt{2\varepsilon_0 P}$$