Explanation:

\(a_1, a_2, ........., a_7, .........., a_{12}, a_{13}\)

Given:
\(a_1, a_2, ........., 10, .........., a_{12}, a_{13}\), and

\(\frac{a_1 + a_2 + ......... + 10 + .......... + a_{12} + a_{13}}{13} = 15\)
i.e.\(a_1 + a_2 + ......... + 10 + .......... + a_{12} + a_{13} = 195\)

Since, we need to find the minimum value of \(a_{13}\), we must keep all other scores maximum
\(10 + 10 + 10 + 10 + 10 + 10 + 10 + a_8 + .......... + a_{12} + a_{13} = 195\)
\(70 + a_8 + a_9 + a_{10} + a_{11} + a_{12} + a_{13} = 195\)
\(a_8 + a_9 + a_{10} + a_{11} + a_{12} + a_{13} = 125\)
\(a_{13} = 125 - (a_8 + a_9 + a_{10} + a_{11} + a_{12})\)

Now, if \(a_8 = a_9 = a_{10} = a_{11} = a_{12} = 20\)
\(a_{13} = 125 - 5(20) = 25\)

if \(a_8 = a_9 = a_{10} = a_{11} = a_{12} = 21\)
\(a_{13} = 125 - 5(21) = 20\), which cannot be the case as \(a_{13}\) has to be the maximum value of all

What if we take \(a_8 = a_9 = a_{10} = a_{11} = 21\) and \(a_{12} = 20\)
\(a_{13} = 125 - 4(21) - 20 = 21\), which again is not possible

Let us take \(a_8 = a_9 = a_{10} = 21\) and \(a_{11} = a_{12} = 20\)
\(a_{13} = 125 - 3(21) - 2(20) = 22\), POSSIBLE

Let us take \(a_8 = a_9 = 21\) and \(a_{10} = a_{11} = a_{12} = 20\)
\(a_{13} = 125 - 2(21) - 3(20) = 23\), the value starts increasing

Col. A: \(22\)
Col. B: \(23\)

Hence, option B