Two blocks possess equal masses. The contact surface of block A...

Let the mass of each block be $m$.

By the principle of conservation of linear momentum during the perfectly inelastic collision, the common velocity $v$ of the blocks after they stick together is given by:
$m u_A + m u_B = (m + m) v$
$m(10) + m(0) = 2mv$
$10m = 2mv \Rightarrow v = 5 \text{ m/s}$

After the collision, the blocks move together as a single system of mass $2m$.
The frictional force acts only on block B, as the surface under block A is smooth.
Frictional force, $f = \mu N_B = \mu mg = 0.10 \times m \times 10 = m$

The deceleration $a$ of the combined system is:
$a = \dfrac{f}{\text{Total mass}} = \dfrac{m}{2m} = 0.5 \text{ m/s}^2$

Using the third equation of motion to find the distance $s$ travelled before coming to rest ($v_f = 0$):
$v_f^2 = v^2 - 2as$
$0 = (5)^2 - 2(0.5)s$
$25 = 1 \cdot s \Rightarrow s = 25 \text{ m}$