How to display modification time of a file?

Don't use ls, this is a job for stat:

stat -c '%y' filename

-c lets us to get specific output, here %y will get us the last modified time of the file in human readable format. To get time in seconds since Epoch use %Y:

stat -c '%Y' filename

If you want the file name too, use %n:

stat -c '%y : %n' filename
stat -c '%Y : %n' filename

Set the format specifiers to suit your need. Check man stat.

Example:

% stat -c '%y' foobar.txt
2016-07-26 12:15:16.897284828 +0600

% stat -c '%Y' foobar.txt
1469513716

% stat -c '%y : %n' foobar.txt
2016-07-26 12:15:16.897284828 +0600 : foobar.txt    

% stat -c '%Y : %n' foobar.txt
1469513716 : foobar.txt

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If you want the output like Tue Jul 26 15:20:59 BST 2016, use the Epoch time as input to date:

% date -d "@$(stat -c '%Y' a.out)" '+%a %b %d %T %Z %Y'
Tue Jul 26 12:15:21 BDT 2016

% date -d "@$(stat -c '%Y' a.out)" '+%c'               
Tue 26 Jul 2016 12:15:21 PM BDT

% date -d "@$(stat -c '%Y' a.out)"
Tue Jul 26 12:15:21 BDT 2016

Check date's format specifiers to meet your need. See man date too.