My electrostatics

No...I had typed the exact question..but I have the answer.

okies.we can find the field if there is a cavity inside the conductor,if the point charge is outside i din find ne clue.we will wait for others.

is dat σq/ε

Nay, mate.

It is σ²2ε₀.

is that\sigma²A/2\epsilon

Its for a unit area...read the question once again.
But, kaise kiya??

Consider a small area element on the conductor. Very close to the surface of that element(outside), it is like a plane(infinite) of charge σ. So, field = σε₀
This filed is due to the contribution of that small area element and the contribution of the rest of the conductor. Consider a point very close to the area element, inside the conductor. Field is zero. This field is also the sum of contributions of the field due to the area element and the rest of the conductor.
But this time the field of the area element is opposite in direction. So the magnitudes of the field due to the area element must be equal to the magnitude of the field due to the rest of the conductor (at the site of the area element)=E. So, for the first case, we can write,
2E=σε₀ or, E=σ2ε₀

Thus the force per unit area on the area element= (field due to the rest of the conductor at the site of the area element)x(charge density)=σ²2ε₀.

bhai unicorn q paper tumhe maalum tha kya ?

this was jee 2010 5 marker direct question

This could have been done just by dimensional analysis.