1. Given data

The numbers are: $3,\;5,\;7,\;2k,\;12,\;16,\;21,\;24$

They are said to be arranged in ascending order, so we must have $7 \le 2k \le 12.$

There are $8$ observations, so the median is the average of the 4th and 5th terms: $\text{Median} = \frac{2k+12}{2} = k+6.$

2. Mean deviation about median

Mean deviation about median is $\frac{1}{8}\sum |x_i-\text{Median}| = 6.$

So, $\sum |x_i-(k+6)| = 48.$

3. Compute each absolute deviation

Since the sequence is ascending and $7 \le 2k \le 12$, the median $k+6$ lies between $2k$ and $12$.

Now compute:

• For $3$: $|3-(k+6)| = k+3$
• For $5$: $|5-(k+6)| = k+1$
• For $7$: $|7-(k+6)| = |1-k|$ Since $2k\ge 7\Rightarrow k\ge 3.5$, so $k>1$, hence $|1-k|=k-1$
• For $2k$: $|2k-(k+6)|=|k-6|$ Since $2k\le 12\Rightarrow k\le 6$, so $|k-6|=6-k$
• For $12$: $|12-(k+6)|=6-k$
• For $16$: $|16-(k+6)|=10-k$
• For $21$: $|21-(k+6)|=15-k$
• For $24$: $|24-(k+6)|=18-k$

4. Add them

Thus, $\sum |x_i-(k+6)| = (k+3)+(k+1)+(k-1)+(6-k)+(6-k)+(10-k)+(15-k)+(18-k).$

Combine constants and coefficients of $k$:

Constants: $3+1-1+6+6+10+15+18=58$

Coefficients of $k$: $k+k+k-k-k-k-k-k = -k$

So, $\sum |x_i-(k+6)| = 58-k.$

Given this equals $48$: $58-k=48$ $k=10.$

5. Find the median

$\text{Median}=k+6=10+6=16.$

But this is impossible, because the data are arranged in ascending order with the 5th term equal to $12$, so the median must lie between the 4th and 5th terms and cannot be $16$.

This contradiction shows our sign assumptions need rechecking. The issue comes from the condition $2k \le 12$, which must hold since the data are already in ascending order. Let us instead use the median directly from the ordered list.

Since there are $8$ terms and the 4th and 5th terms are $2k$ and $12$, the median is $M=\frac{2k+12}{2}=k+6.$

Now because $7\le 2k\le 12$, we get $3.5\le k\le 6,$ so indeed all sign choices above were valid. Hence the equation $58-k=48$ Rightarrow k=10 is inconsistent with the ordering condition.

Therefore, the given question data are inconsistent as stated.

6. Check options directly

The possible median is $k+6$, with $3.5\le k\le 6$, so $9.5\le \text{Median}\le 12.$ Among the options, possible values are:

• $10.5$
• $11$
• $11.5$
• $12$

Let us test option $11$: $k+6=11 \Rightarrow k=5 \Rightarrow 2k=10.$ Then data are $3,5,7,10,12,16,21,24.$

Median $=11$.

Mean deviation about median: $\frac{|3-11|+|5-11|+|7-11|+|10-11|+|12-11|+|16-11|+|21-11|+|24-11|}{8}$ $=\frac{8+6+4+1+1+5+10+13}{8}=\frac{48}{8}=6.$

This satisfies the condition exactly.

Hence the median is $\boxed{11}.$