Time for serial data(UART) transmission
The number of bits to transmit is 32. In addition, there is 1 start and 1 stop bit per 8 or 9 bits (depending what is selected for #data bits), assuming 8 data bits, there are 4 start bits and 4 stop bits, so in total 32 + 4 + 4 = 40 bits.

115200 baud means 11500 bits/sec, so 40 bits will take 40 / 115200 = appr. 0.000347 s = 347 us.

However, there might be a slight overhead / delay for the processing itself, but this is high likely negligible.

UPDATE

The question seems to have been edited. The above is true for sending 4 bytes (32 bits) of real data (resulting in 40 bits). In the example above 6 bytes are sent, but the formula stays the same.