The minimum energy required to launch a m kg satellite from earths surface in a circular orbit at an atitude of 2R , where R is the radius of earth .

The minimum energy should be (1/2) mv$^2$

As it is circular orbit

$$mv^2/(3R) = GMm/(3R)^2$

Solving this I got

$$(1/2)mv^2 = GMm/(6R) = mgR/6$

But the answer is given as (5/6)mgR

Try applying Conservation of Energy.