If you see the content of master information block, you can easily calculate the size of master information block. Total size of content of MIB is 24 bits. Calculation details is as following:
3 bits for downlink bandwidth
3 bits for phich-config
8 bits for System Frame Number
10 bits as spare
These 24 bits are not sent directly over PBCH. There are lot of things done to send over PBCH.
16 bits long CRC is generated and scrambled with antenna specific mask.
Now total 40 bits (24 +16) is passed to convolution encoding which generates 3 streams of 40 bits. Now 120 bits are passed to rate matching function which just repeats these 120 bits 16 times and information becomes 1920 bits long.
QPSK is used to transmit the MIB information that's why 1920 bits information becomes 960 QPSK symbols.
These 940 QPSK symbols should be transmitted within 40ms since a new MIB information can come to broadcast. 960 / 4 = 240 QPSK symbols should be transmitted in each radio frame. For that purpose 240 Resource Elements are required in subframe zero of each radio frame since MIB is transmitted in slot 1 of subframe zero. Since in each Radio block, 8 REs are used for reference signals, so only 40 REs will be available per RB to carry MIBs information. Because there are 240 bits so you need 240/40 = 6 RBs to transmit MIB information. However total 960 QPSK symbols are transmitted by using 4 radio frames.
Hope, this detailed information will help.