Let the first odd number be 'x'

Then, Second odd number is 'y'

Then, x+y=38    ---Eq(1)

And   x×y=325

(x-y)²=(x+y)²-4xy

(x-y)²⇒ (38)²-4×325

(x-y)²⇒ 1444-1300

x-y⇒ √144

or x-y=12         ---Eq(2)

By solving Eq(1) and Eq(2), we get

⇒ 2x=50

⇒ x=25, y=13

∴ Required number =3×25=75

Hence, Option B is correct.