It certainly would be nice to know the target RDBMS. But this question is asked so often so let's try and list'em all (at least popular ones) side by side.
For SQL Server:
SELECT e.Ename, e.Eage, e.Eadd, e.Ephone, d.dname
FROM Employee e LEFT JOIN
(
SELECT fkEmpid,
STUFF((SELECT ',' + dname
FROM Department
WHERE fkEmpid = t.fkEmpid
FOR XML PATH('')) , 1 , 1 , '' ) dname
FROM Department t
GROUP BY fkEmpid
) d
ON e.Empid = d.fkEmpid
Here is SQLFiddle demo
For Mysql, SQLite, HSQLDB 2.X:
SELECT e.Ename, e.Eage, e.Eadd, e.Ephone, d.dname
FROM Employee e LEFT JOIN
(
SELECT fkEmpid,
GROUP_CONCAT(dname) dname
FROM Department t
GROUP BY fkEmpid
) d
ON e.Empid = d.fkEmpid
Here is SQLFiddle demo (MySql)
Here is SQLFiddle demo (SQLite)
For Oracle 11g:
SELECT e.Ename, e.Eage, e.Eadd, e.Ephone, d.dname
FROM Employee e LEFT JOIN
(
SELECT fkEmpid,
LISTAGG (dname, ',') WITHIN GROUP (ORDER BY dname) dname
FROM Department t
GROUP BY fkEmpid
) d
ON e.Empid = d.fkEmpid
Here is SQLFiddle demo
For PostgreSQL 9.X:
SELECT e.Ename, e.Eage, e.Eadd, e.Ephone, d.dname
FROM Employee e LEFT JOIN
(
SELECT fkEmpid,
string_agg(dname, ',') dname
FROM Department t
GROUP BY fkEmpid
) d
ON e.Empid = d.fkEmpid
Here is SQLFiddle demo
Output in all cases:
| ENAME | EAGE | EADD | EPHONE | DNAME |
---------------------------------------------
| x | 23 | b | 677 | test,test1 |
| y | 24 | h | 809 | hello |
| z | 34 | u | 799 | (null) |