Given data

• Mass of block: $m=1\,\text{kg}$
• Inclination angle: $\theta=60^\circ$
• Applied force along incline: $F=10\,\text{N}$
• Displacement along incline: $s=10\,\text{m}$
• Acceleration due to gravity: $g=10\,\text{m/s}^2$

Forces along the incline Along the inclined plane:

• Upward force: applied force $F=10\,\text{N}$
• Downward component of weight: $mg\sin\theta = 1\cdot 10\cdot \sin 60^\circ = 10\cdot \frac{\sqrt3}{2}=5\sqrt3\,\text{N}$
• Friction acts downward along the plane.

Interpretation of the question Since the options are small and the phrase asks for the work done against frictional force, we take the motion to be at constant speed / no change in kinetic energy, so net force along incline is zero.

Therefore, $F = mg\sin\theta + f$ where $f$ is the frictional force.

So, $f = F - mg\sin\theta = 10 - 5\sqrt3$

Work done against friction Work done against friction over distance $s=10\,\text{m}$ is $W_f = f\,s = (10-5\sqrt3)\times 10$ $W_f = 100 - 50\sqrt3$

Using $\sqrt3\approx 1.732$, $W_f \approx 100 - 86.6 = 13.4\,\text{J}$

Comparison with options The computed value $13.4\,\text{J}$ does not match any option exactly. The nearest/simple intended value among options is likely $10\,\text{J}$ (Option D) only if the question/data has a typo (for example, displacement or angle).

With the given data, the exact answer is: $\boxed{100-50\sqrt3\,\text{J}}$

Conclusion Hence the stored answer B (5 J) is not consistent with the given values.