Let n and k be positive integers and σ(n) the sum of all positive divisors of n. We call n an exactly k-deficient-perfect number with deficient divisors d1, d2, . . . , d k if d1, d2, . . . , d k are distinct proper divisors of n and σ(n) = 2n − (d1 + d2 + . . . + d k ). In this article, we show that the only odd exactly 3-deficient-perfect number with at most two distinct prime factors is 1521 = 3 2• 13 2 .