Yay For 2014!

From the lower powers of $2014$ , we notice that $m$ of $2014^n \equiv m \pmod {100}$ only depends on the last digit of $n$ for $n \ge 2$ . If $n \equiv d_0 \pmod {10}$ , then $m$ are as follows:

$\begin{matrix} { d }_{ 0 } & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ m & 76 & 64 & 96 & 44 & 16 & 24 & 36 & 4 & 56 & 84 \end{matrix}$

Since $2014 \equiv 4 \pmod {10} \Rightarrow 2014^{2014} \equiv 16 \pmod {100} \equiv 6 \pmod {10}$

$\Rightarrow 2014^{2014^{2014}} \equiv 2014^6 \pmod {100} \equiv 36 \pmod {100}$

Therefore its tens digit is $\boxed {3}$ .

$2014^{2014^{2014}} \equiv 14^{2014^{2014}}(mod\:100)$

As Utkarsh Dwivedi rightfully said, the last two digits of the powers of 14 follow a pattern: $14^n \equiv 14^{n+10} (mod\:100) \forall n\in\mathbb{N} , n>1$

It follows that the required answer depends on $2014^{2014} (mod\:10)$ . This is no different from $4^{2014} (mod\:10)$ . Note that the powers of 4(mod 10) cycle between 4 and 6, depending on whether the exponent is odd or even, respectively.

2014 is even $\implies 2014^{2014} \equiv 6 (mod\:10) \implies 2014^{2014^{2014}} \equiv 36 (mod\:100)$ , the tens digit of which is $\boxed{3}$

Let us take, for sake of convenience $p=2014^{2014^{2014}}$ . Now we know in $p$ , actually while computing its value the tens and the units digits would not be affected by the hundreds, thousands or so on. Instead it is clearly evident that the tens and units digit of $p$ would be same as $14^{2014^{2014}}$ . We find that the powers of 14 have the tens and units digits as $14,96,44,16,24,36,04,56,84,76,64,96,44.....$ starting from power 1 to endless . So, we find 96 will repeat itself in the digits after every 10 powers after the first power of 14.So we just need to find the remainder when $2014^{2014}-1$ (because the first power digit of 14 is 14 itself and it does not repeat itself) is divided by 10. It's 5 and I think everyone knows how to find it. Since the fifth power digit is 36. So the tens digit is $\boxed{3}$ .This is how I solved but there may be much better methods.

A systematic, unimaginative, brute-forcey Chinese Remainder solution:

Let $n = 2014, q = n^n, p = n^q$ . Then we wish to find the class of $p$ $\mod 100$ , which I will write ${[p]}_{100}$ .

Obviously $p \equiv 0 \pmod 4$ , so we now need ${[p]}_{25} = {[n^q]}_{25}$ . Noticing that $\phi(25) = 20$ , it is sufficient (by Euler's theorem) to find ${[q]}_{20}$ , which can again be separated into $q \equiv 0 \pmod 4$ and $q \equiv n^2 \pmod 5 \equiv 4^2 \pmod 5$ . This gives us $16$ , a simultaneous solution of both these equations, and hence $q \equiv 16 \pmod {20}$ .

We go back up, to $25$ , where this gives us $p \equiv n^{16} \equiv 14^{16} \equiv 196^8 \equiv {(-4)}^8 \equiv 256^2 \equiv 36$ modulo $25$ . Notice that $36\equiv p\mod 4$ , so $p \equiv 36 \pmod {100}$ and the answer is $3$ .

the tenth place digits will be coming as 9,1,3,5,7 for powers of 2,4,6,8,10 and for for 12 same value of 2 in a continuous manner and for this we get the value of power 6, so digit is 3