We need to find the units digit of \((24)^{5+2x}\) + \((36)^6\) + \((17)^3\)

Now, Units digit of \((24)^{5+2x}\) will be same as units digit of \((4)^{5+2x}\) (As units digit of 24 is same as 4 which is 4)
Units digit of \((36)^6\) will be same as units digit of \((6)^6\) (As units digit of 36 is same as 6 which is 6)
Units digit of \((17)^3\) will be same as units digit of \((7)^3\) (As units digit of 17 is same as 7 which is 7)

Let's find the units digit of \((4)^{5+2x}\)

To find the units digit of power of 4 we need to check the cyclicity in the units digit of powers of 4

\(4^1\) units’ digit is 4 [ 4 ]
\(4^2\) units’ digit is 6 [ 16 ]
\(4^3\) units’ digit is 4 [ 64 ]
\(4^4\) units’ digit is 6 [ 256 ]

=> Units digit of any positive odd integer power of 4 is 4
=> Units digit of any positive even integer power of 4 is 6
=> Now 2x + 5 will be Evan + Odd = Odd (As 2x is Even and 5 is odd)
=> Units digit of \((4)^{5+2x}\) = 4

Let's find the units digit of \((6)^6\)

To find the units digit of power of 6 we need to check the cyclicity in the units digit of powers of 6

\(6^1\) units’ digit is 6 [ 6 ]
\(6^2\) units’ digit is 3 [ 36 ]
\(6^3\) units’ digit is 3 [ 216 ]

=> Units digit of any positive integer power of 6 is 6
=> The units digit of \(6^6\) = 6

Let's find the units digit of \((7)^3\)

\(7^1\) units’ digit is 7 [ 7 ]
\(7^2\) units’ digit is 9 [ 7*7 = 49 ]
\(7^3\) units’ digit is 3 [ 9*7 = 63 ] (P.S>: we don't need to proceed further)
\(7^4\) units’ digit is 1 [ 3*7 = 21 ]
\(7^5\) units’ digit is 3 [ 1*7 = 7 ]

=> Units digit of power of 7 repeats after every \(4^th\) number

=> Units digit of \((24)^{5+2x}\) + \((36)^6\) + \((17)^3\) = Units digit of 4 + 6 + 3 = Units digit of 13 = 3

So, Answer will be B.
Hope it helps!

Watch the following video (from 2:48 mins) to learn how to find cyclicity of 3 and other numbers