How to get semi-major axis from TLE?

This should be fairly straightforward, but, how do I get the semi-major axis (a) from at TLE.

For example if I have the TLE:

1 25544U 98067A 01260.91843750 .00059354 00000-0 74277-3 0 4795
2 25544 51.6396 342.1053 0008148 106.9025 231.8021 15.5918272116154

I know that 15.5918272 is the mean motion of the body ($n$). I also know that $n = \sqrt{\frac{\mu}{a^3}}$. If I use the given $n$ value I get a semi-major axis $a=11.79$ which is obviously incorrect. What am I missing here?

The TLE gives mean motion ($n$) in $\frac{rev}{day}$. This needs to be converted to $\frac{rad}{s}$ which can be accomplished by multiplying the $n$ TLE value by $\frac{2\pi}{86400}$.
Therefore, to go directly from $n$ in TLE to the semi-major axis $a$. We can use the following formula: $a=\frac{u^{1/3}}{\frac{2n\pi}{86400}^{2/3}}$.
For $n=15.5918272 \,\, \frac{rev}{day}$, we get $a=6768.16 \,\, km$.
• In this answer, $u$ refers here to $\mu$, the Standard gravitational parameter ($3.986004418 ×10^{14} m^3 s^{−2}$ for the Earth) – Covich Oct 10 '17 at 15:16