The sum of the first \(n\) terms of an arithmetic progression is given by \(S_n = \frac{n}{2} (2a + (n-1)d)\).

Equating this to \(n^2 + 4n\), we have:

\(\frac{n}{2} (2a + (n-1)d) = n^2 + 4n\).

Comparing coefficients, we get:

\(\frac{d}{2} = 1\) and \(a - \frac{1}{2}d = 4\).

Solving these, \(d = 2\) and \(a = 5\).

The \(k\)th term is given by \(a + (k-1)d\).

Substituting the values of \(a\) and \(d\), we have:

\(5 + 2(k-1) > 200\).

Simplifying, \(2k + 3 > 200\).

Solving for \(k\), we get \(k > 98.5\).

Thus, the smallest integer value of \(k\) is \(99\).