Interpret the statement

The phrase "frictionless force on the block" means the force of friction acting on the block is $10\,\text{N}$.

Since the block is at rest on a rough incline, static friction balances the component of weight along the plane.

Resolve weight along the incline

For a block of mass $m$ on an incline of angle $30^\circ$, $\text{component of weight along plane} = mg\sin 30^\circ$

Given: $g=10\,\text{m/s}^2, \quad \sin 30^\circ = \frac{1}{2}$

So, $mg\sin 30^\circ = m\cdot 10 \cdot \frac{1}{2} = 5m$

Use equilibrium condition

Since the block is at rest, static friction must equal this component: $f = mg\sin 30^\circ$

Given friction force, $f=10\,\text{N}$

Therefore, $5m=10$ $m=2\,\text{kg}$

Check static friction limit

Maximum static friction is: $f_{s,\max}=\mu_s N$ where $N=mg\cos 30^\circ = 2\cdot 10 \cdot \frac{\sqrt{3}}{2}=10\sqrt{3}$

Hence, $f_{s,\max}=0.8\times 10\sqrt{3}=8\sqrt{3}\approx 13.9\,\text{N}$

Since required friction is $10\,\text{N}$ and $10 < 13.9,$ static equilibrium is possible.

Match with options

$m=2.0\,\text{kg}$

So the correct option is C.