1. Write the data set

The numbers are: $a, 2a, 3a, \dots, 50a$

So there are $50$ observations.

2. Find the median

For $50$ terms, the median is the average of the $25^{\text{th}}$ and $26^{\text{th}}$ terms.

These are: $25a \quad \text{and} \quad 26a$

Hence median $M$ is $M=\frac{25a+26a}{2}=\frac{51a}{2}$

3. Mean deviation about the median

Mean deviation about median is $\text{MD} = \frac{1}{50}\sum_{k=1}^{50}\left|ka-\frac{51a}{2}\right|$

Factor out $|a|$: $\text{MD} = \frac{|a|}{50}\sum_{k=1}^{50}\left|k-\frac{51}{2}\right|$

Given that this mean deviation is $50$, so $\frac{|a|}{50}\sum_{k=1}^{50}\left|k-\frac{51}{2}\right|=50$

4. Evaluate the sum

We need $\sum_{k=1}^{50}\left|k-\frac{51}{2}\right|$

Since $\frac{51}{2}=25.5$, the distances are: $24.5, 23.5, 22.5, \dots, 0.5, 0.5, \dots, 22.5, 23.5, 24.5$

Thus, $\sum_{k=1}^{50}\left|k-25.5\right| = 2(0.5+1.5+2.5+\dots+24.5)$

Now, $0.5+1.5+2.5+\dots+24.5$ is an AP with $25$ terms.

Its sum is $\frac{25}{2}(0.5+24.5)=\frac{25}{2}(25)=\frac{625}{2}$

Therefore, $\sum_{k=1}^{50}\left|k-25.5\right| = 2\cdot \frac{625}{2}=625$

5. Use the given mean deviation

So, $\frac{|a|}{50}\cdot 625=50$

$|a|\cdot \frac{625}{50}=50$

$|a|\cdot 12.5=50$

$|a|=\frac{50}{12.5}=4$

6. Match with options

Thus, $|a|=4$

So the correct option is A.