The coefficient of friction between the table and the block shown...

Let $a$ be the acceleration of the system and $T_1, T_2$ be the tensions in the left and right strings respectively. Taking $g = 10 \text{ m/s}^2$.

The $15 \text{ kg}$ block moves downwards, the $5 \text{ kg}$ block on the table moves to the left, and the $5 \text{ kg}$ hanging block moves upwards.

The kinetic friction on the $5 \text{ kg}$ block on the table is:
$f = \mu N = \mu m g = 0.2 \times 5 \times 10 = 10 \text{ N}$

The equations of motion for the three blocks are:
For the $15 \text{ kg}$ block: $15g - T_1 = 15a \Rightarrow 150 - T_1 = 15a$
For the $5 \text{ kg}$ block on the table: $T_1 - T_2 - f = 5a \Rightarrow T_1 - T_2 - 10 = 5a$
For the $5 \text{ kg}$ hanging block: $T_2 - 5g = 5a \Rightarrow T_2 - 50 = 5a$

Adding the three equations:
$(150 - T_1) + (T_1 - T_2 - 10) + (T_2 - 50) = 15a + 5a + 5a$
$90 = 25a \Rightarrow a = 3.6 \text{ m/s}^2$

Substituting $a$ into the first equation:
$150 - T_1 = 15 \times 3.6 = 54 \Rightarrow T_1 = 96 \text{ N}$

Substituting $a$ into the third equation:
$T_2 - 50 = 5 \times 3.6 = 18 \Rightarrow T_2 = 68 \text{ N}$

The tension in the left string is $96 \text{ N}$ and in the right string is $68 \text{ N}$.