That looks like homework for school. We don't help with that.
Moreover, it's an unanswerable question, because it doesn't include the processing time in the browser, the OS, antivirus and the videodriver and the fact that it makes a difference if the browser asks ask for the 3 pictures in parallel or not.
To see a real life example of processing in a browser visit any webpage after (in Firefox) pressing Tools>Webdeveloper>Network.
In your formula, I don't see the size of the webpage and the three images. So your answer certainly is wrong. Big pictures take more time than small pictures.
I'm having trouble understanding this question. Can anyone help?......
Suppose within your Web browser you click on a link to obtain a Web page. The content of the web page is not cached in the local caching server, and the IP address for the associated URL is not cached in your local host either, so that a DNS look-up is necessary to obtain the IP address. Suppose that n DNS servers need to be visited before your host receives the IP address from DNS, and each of successive visits incur round trip delay of RTT1, RTT2 ... RTTn respectively. Further suppose that three images, sized 100KB, 120KB and 80KB respectively, are embedded in the Web page associated with the link, which are stored on the same web server as the web page itself, while the size of the Web page is 4KB. Let RTT0 denote the round trip delay between your computer and the server containing the web page, and let R denote the average end-to-end transmission rate between your computer and the Web server. How much time elapses from when you click on the link until you see the entire content of the web page associated with the link? Your calculation should include all the RTTs and transmission delays and show them in a properly simplified mathematical formula.
Would something like this make sense?
T = (2RTT0 / R) + (2RTT1 / R) + (2RTT2 / R) + (2RTT3 / R)
That's 2 RTT's for the webpage and 2 more for each of the 3 objects, each divided by the transmission rate?