Method $1$: Long$-$Hand $($Manual Step$-$by$-$Step$)$

Step $1$: Convert Binary to Decimal

Converting $110($binary$)$ to Decimal

$1102=(122)+(121)+(020)110\_2=(1\backslash$times $2^2)+(1\backslash$times $2^1)+(0\backslash$times $2^0)1102=(122)+(121)+(020)=4+2+0=610=4+2+0=6\_\{10\}=4+2+0=610$

Converting $110101($binary$)$ to Decimal

$1101012=(125)+(124)+(023)+(122)+(021)+(120)110101\_2=(1\backslash$times $2^5)+(1\backslash$times $2^4)+(0\backslash$times $2^3)+(1\backslash$times $2^2)+(0\backslash$times $2^1)+(1\backslash$times $2^0)1101012=(125)+(124)+(023)+(122)+(021)+(120)=32+16+0+4+0+1=5310=32+16+0+4+0+1=53\_\{10\}=32+16+0+4+0+1=5310$

Step $2$: Convert Decimal to Hexadecimal

Converting $6($decimal$)$ to Hexadecimal

$6106\_\{10\}610$ is already less than $16,$ so it directly converts to $6$ in hexadecimal.

$610=6166\_\{10\}=6\_\{16\}610=616$

Converting $53($decimal$)$ to Hexadecimal

Divide $53$ by $16$ and record the quotient and remainder.

$5316=3($quotient$),5($remainder$)53\backslash$div $16=3\backslash$quad $\backslash$text$\{($quotient$)\},\backslash$quad $5\backslash$quad $\backslash$text$\{($remainder$)\}5316=3($quotient$),5($remainder$)$

The hexadecimal equivalent is obtained by reading the remainder and quotient in order:

$5310=351653\_\{10\}=35\_\{16\}5310=3516$