How can I avoid calculating with huge numbers when implementing the RSA algorithm There is 26-letter English alphabet. 

There is the plain text: TRYAGAINLATER.
I need to encrypt it by RSA algorithm with the public key 53.
What is the ciphertext?

Therefore:
M = TRYAGAINLATER,
M=19 17 24 0 6 0 8 13 11 0 19 4 17.

Encryption: $c = m^e \pmod n = m^{53} \pmod{26}$

The problem is that $m^{53}$ is huge number. How can I avoid this?
$\def\Z{\mathbb Z}$Usual (i.e. real life) RSA works with huge numbers, not with small numbers like 53 and 26. (The numbers there are, for example, 1024-bit numbers for the lowest acceptable security level nowadays.)

But anyway, the core mathematic operation behind RSA is modular exponentiation - you exponentiate not in the ring of integers $\Z$, but in the modular ring $\Z/_{n&#183;\Z}$ (where $n$ is the modulus). You get the same result by exponentiating in $\Z$ and then reducing, or by reducing after each multiplication step ... and the latter avoids the overhead of really huge numbers.

So, you calculate $c = (…((m &#183; m \bmod n) &#183; m \bmod n ) &#183; … &#183; m \bmod n)$, such that each intermediary result is a number smaller than $n$, instead of calculating the huge number $m^e$ directly.

(This is the naive exponentiation, which takes $e$ steps - there are faster algorithms which only need around $\log_2 e$ steps, like square-and-multiply.)

But your example is faulty, too - the public exponent is normally a number smaller than the modulus (which is a part of the public key, too, and actually the more important part).