[uuDevil]

I'm unreliable, but my answer is 25.

Method:

The expected #failures in a 3-hr period is 400/(24*365/3)=.137

The probability of 2 or more failures in a 3-hr period is

1-P(X=0)-P(X=1)= 1-exp(-.137)*(.137)^0/0!-exp(-.137)*(.137)^1/1!= .0857

The expected number of times 2 or more failures will occur in the same 3-hr period over a year is

.0857*(24*365/3)=25.0