In this solution we have considered 8 bits representation which can be extended up to 32 bits using the same idea.

    Let first no whose bit is to be replaced is:
    7 //0000 0111

    Second no from whom bit is to be replaced
    8 //0000 1000

    Specified position: 3 (0-indxed)

First no: 7

Second no: 8

Algorithm:

IF specified bit at second no is 1
    1)  Form a bit mask (0, d)
        0 - for all position except the bit specified
        d - 1 at specified position(dth)
    2)  Do bitwise OR between the first no and the mask 
ELSE //specified bit at second no 0
    1)  Form a bit mask (1, d)
        1 - For all position except the bit specified
        d - 0 at specified position (dth)
    2)  Do bitwise AND between the first no & the mask.

Forming the bitmask (0, d)

Let the specified position d.
Declare temp as (second>>d) & 1
Right shift second number d times to get the dth bit as LSB
Bitwise AND with 1
This results in 0000 000di(where diis the dth bit of second no)
Left shift temp d times. This results in our desired bitmask(0, d)

Forming the bitmask (1, d)
Declare flag 255;//FF, all bit set to 1(considering 8 bit)
Declare temp as 1<<d;
Do bitwise XOR between flag and temp
//this sets only the specified position bit 0 and others with 1
This results in our desired bitmask (1, d)

C implementation

#include <stdio.h>

int main()
{
	int first,second,pos;

	printf("enter first & second no:\n");
	scanf("%d %d",&first,&second);

	printf("enter specified position(0-indexed)\n");
	scanf("%d",&pos);

	//collect corresponding bit of second no
	int temp=(second>>pos)&1;

	//if bit at specified position is 1
	if(temp==1){
		temp=temp<<pos;
		first|=temp;
	}
	else{ //if bit at specified position is 0
		int flag=255;//FF, all bit set to 1(considering 8 bit)
		temp=1<<pos;
		//this set only the specified position bit 0 others 1
		flag=flag^temp;
		first&=flag;
	}

	printf("converted no %d\n",first);

	return 0;
}

Output

First run:
enter first & second no:
7 8
enter specified position(0-indexed)
3
converted no 15

Second run:
enter first & second no:
7 8
enter specified position(0-indexed)
2
converted no 3