Step 1: Understand pulley system
The small pulley is attached to the block → it is a movable pulley

So, the block is pulled by two tensions:
Total pull on block = $2T$

Step 2: Forces on block (along incline)

Weight component down the plane:
$ 50g \sin 30^\circ = 50g \times \frac{1}{2} = 25g $

Normal reaction:
$ N = 50g \cos 30^\circ = 50g \times \frac{\sqrt{3}}{2} $

Friction (opposing upward motion):
$ f = \mu N = 0.3 \times 50g \times \frac{\sqrt{3}}{2} $

Step 3: Condition for upward motion

$ 2T = 25g + 0.3 \times 50g \times \frac{\sqrt{3}}{2} $

Simplify:
$ 2T = 25g + 15g \times \frac{\sqrt{3}}{2} $

Step 4: Relation with hanging mass

$ T = Mg $

So:
$ 2Mg = 25g + \frac{15\sqrt{3}}{2}g $

Cancel $g$:
$ 2M = 25 + \frac{15\sqrt{3}}{2} $

Step 5: Solve for $M$

$ M = \frac{1}{2}\left(25 + \frac{15\sqrt{3}}{2}\right) $

$ \sqrt{3} \approx 1.732 $

$ M = \frac{1}{2}(25 + 12.99) = \frac{37.99}{2} \approx 19.0 $

Final Answer:
$\boxed{19.0 \text{ kg}}$