1. Given data

• Mass of block: $m=25\,\text{kg}$
• Coefficient of friction: $\mu=0.25$
• Force is applied at angle: $\theta=45^\circ$
• Displacement: $s=5\,\text{m}$
• Block moves with uniform velocity $\Rightarrow$ net force is zero.

2. Forces on the block

Let the applied force be $F$.

Its components are:

$F_x=F\cos45^\circ, \qquad F_y=F\sin45^\circ$

Since the force is pulling upward at $45^\circ$, the normal reaction becomes:

$N=mg-F\sin45^\circ$

Friction is:

$f=\mu N=\mu\left(mg-F\sin45^\circ\right)$

3. Horizontal force balance

Because velocity is uniform,

$F\cos45^\circ=f$

So,

$F\cos45^\circ=\mu\left(mg-F\sin45^\circ\right)$

Substitute $\mu=0.25$ and $\sin45^\circ=\cos45^\circ=\frac{1}{\sqrt{2}}$:

$F\frac{1}{\sqrt{2}}=0.25\left(25g-F\frac{1}{\sqrt{2}}\right)$

Using $g=9.8\,\text{m/s}^2$,

$F\frac{1}{\sqrt{2}}=0.25\left(245-F\frac{1}{\sqrt{2}}\right)$

$F\frac{1}{\sqrt{2}}=61.25-0.25F\frac{1}{\sqrt{2}}$

$1.25F\frac{1}{\sqrt{2}}=61.25$

$F\frac{1}{\sqrt{2}}=49\,\text{N}$

Thus,

$F\cos45^\circ=49\,\text{N}$

4. Work done by the applied force

The displacement is horizontal, so only the horizontal component of the force does work:

$W=(F\cos45^\circ)s$

$W=49\times 5=245\,\text{J}$

5. Correct option

$\boxed{245\,\text{J}}$

So the correct option is D.