If the mean deviation about the median of the numbers $$a, 2a, \ldots, 50a$$ is $$50$$, then $$|a|$$ equals:

The observations are $$a,\,2a,\,3a,\,\ldots,\,50a$$, so the total number of terms is $$n = 50$$.

Median
Because $$n$$ is even, the median $$M$$ is the average of the $$25^{\text{th}}$$ and $$26^{\text{th}}$$ terms when the data are arranged in ascending order.

If $$a \gt 0$$, the sequence is already increasing; the two central terms are $$25a$$ and $$26a$$.
If $$a \lt 0$$, the sequence decreases, but after arranging it in ascending order the two central terms again turn out to be $$26a$$ and $$25a$$ (just reversed).
Thus in either case

$$M = \frac{25a + 26a}{2} = 25.5\,a$$.

Mean deviation about the median
Definition: $$\text{M.D.} = \frac{1}{n}\sum_{i=1}^{n}\lvert x_i - M\rvert$$.

Factorising $$a$$ from each absolute difference,

$$\sum_{k=1}^{50}\lvert ka - 25.5a\rvert = |a|\sum_{k=1}^{50}\lvert k - 25.5\rvert.$$

The inner sum is purely numerical. Because of symmetry about $$25.5$$,

$$\sum_{k=1}^{50}\lvert k - 25.5\rvert = 2\sum_{k=1}^{25}(25.5 - k).$$

Compute that finite arithmetic series:
$$\sum_{k=1}^{25}(25.5 - k) = 25\cdot25.5 \;-\;\frac{25\cdot26}{2} = 637.5 - 325 = 312.5.$$

Hence the total absolute deviation is
$$2 \times 312.5 = 625.$$

Therefore

$$\text{M.D.} = \frac{1}{50}\bigl(|a|\times625\bigr) = 12.5\,|a|.$$

Given that the mean deviation about the median equals $$50$$,

$$12.5\,|a| = 50 \;\Longrightarrow\; |a| = \frac{50}{12.5} = 4.$$

Hence $$|a| = 4$$.

Option B which is: $$4$$