Select any digit from $10$ digits, one uppercase alphabet from $26$ alphabets and one lowercase alphabets from $26$ alphabet, each combination can permute $3!$ times for ex: $5Ab, 5bA, A5b, Ab5, bA5, b5A$
So, total possible passwords= $(10C1 * 26C1 * 26C1 ) * 3! $
$= 40560$
Hence option c) is correct