Only One

Since each palindrome with an even number of digits is divisible by $11$ - the only palindromic prime with an even number of digits is $11$ .

Every palindromic number with an even number of digits is divisible by 11. 11 is only divisible by 1 and itself therefore 11 is both palindromic, prime and has an even number of digits.

The smallest prime palindromic numbers are: 2, 3, 5, 7, 11, 101, 131, 151, 181, 191, 313, 353, 373, 383, 727, 757, 787, 797, 919, 929, 10301, 10501, 10601, 11311, 11411, 12421, 12721, 12821, 13331, 13831, 13931, 14341, 14741, 15451, 15551, 16061, 16361, 16561, 16661, 17471, 17971, 18181, 18481, 19391, 19891, 19991, ...

The only that satisfy the requirements is: 11; Then, the answer is $\boxed {11}$

A palindrome is a number that reads the same from left to right and from right to left

For example, 1331, 202 ,12345654321,33,121

Further, a prime is a number which has only two factors ie. 1 and itself.

A palindromic prime (sometimes called a palprime) is a prime number that is also a palindromic number. Palindromicity depends on the base of the numbering system and its writing conventions, while primality is independent of such concerns. The first few decimal palindromic primes are: 2, 3, 5, 7, 11, 101, 131, 151, 181, 191, 313, 353, 373, 383, 727, 757, 787, 797, 919, 929, …

Except for 11, all palindromic primes have an odd number of digits, because the divisibility test for 11 tells us that every palindromic number with an even number of digits is a multiple of 11.

Thus, 11 is the answer to the question

Every palindromic number with an even number of digits is divisible by 11 . And we know 11 is only divisible by 1 and itself therefore 11 is both palindromic, prime and has an even number of digits...... simple is it...

11 is a prime number with even number of digits.