I'm not sure I understand your question, but I suspect the answer is that you need to use double quotes (") to interpolate the contents of $data into the string you are building. I'd do it like:

my $data_brf = "$data_dir\\$data\.brf";
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Basically I want to append "$data.brf" to "$data_dir" and then print it,so am I am trying to interpolate $data,if I use the way you suggested..I am not able to append...

$data_brf=$data_dir.'\\$data\.brf';
print "$data_brf";
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I suggest you copy and paste the code I gave earlier. It is not what you implied I suggested. Consider:

use strict;
use warnings;

my $data_dir = 'c:\\Wibble';
my $data = 'Plonk';
my $data_brf = "$data_dir\\$data\.brf";

print "$data_brf\n";
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Prints:

c:\Wibble\Plonk.brf
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If that is not what you want you better show us the contents of $data_dir and $data, and show us what you expect to see printed.

True laziness is hard work

I am trying to decode the value in "$data"

What you're trying to do is interpolate $data, or append it to $data_brf

Since single quotes do not interpolate, you get a literal $data instead of the contents of $data (interpolation)

As perlintro tells us, double quotes interpolate, so use double quotes

 $data_brf = "$data_dir\\$data\.brf";

 $data_brf = qq{$data_dir\\$data\.brf};

 $data_brf = qq[$data_dir\\$data\.brf];

 $data_brf = qq<$data_dir\\$data\.brf>;

 $data_brf = qq($data_dir\\$data\.brf);

 $data_brf = qq'$data_dir\\$data\.brf';

My goal is I want to append "$data.brf" to "$data_dir" and then print it,so am I am trying to interpolate $data,if I use the way you suggested..I am not able to append...

$data_brf=$data_dir.'\\$data\.brf';
print "$data_brf";
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I suggested you use double-quotes, I showed you six different ways, simply copy and paste one of them