We have to send 800 + 20 = 820 bytes.
From P to R1:
Maximum frame length supported is 1024 and 820 + 12 is within one frame- 832 bytes transferred.
From R1 to R2:
Maximum frame length is 256 bytes including 8 bytes header. We have 832 - 12 = 820 bytes (IP header added by P would be removed here). So, this requires ceil (820/(256 - 8)) = 4 frames - 3 * 256 + 820 - (248*3) + 8 = 852 bytes transferred.
From R2 to Q:
We need to send 248 bytes in a frame and maximum frame length is 512 bytes including 12 bytes header. So, four frames frames are required - 3 * (248 + 12) + (76 + 12) = 868 bytes transferred.